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Proof step 2 from proof in-05 (previous step)

s ++ a = s0 ++ b ⊦ a = b, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a0. a0 = a ∨ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a = b, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s;